Invert actual bits of a number
Given a non-negative integer n. The problem is to invert the bits of n and print the number obtained after inverting the bits. Note that the actual binary representation of the number is being considered for inverting the bits, no leading 0’s are being considered.
Examples:
Input : 11
Output : 4
(11)10 = (1011)[2]
After inverting the bits, we get:
(0100)2 = (4)10.
Input : 10
Output : 5
(10)10 = (1010)2.
After reversing the bits we get:
(0101)2 = (101)2
= (5)10.
Method 1 (Using bitwise operators)
Prerequisite: Toggling k-th bit of a number
C++
#include <bits/stdc++.h>
using namespace std;
void invertBits( int num)
{
int x = log2(num) + 1;
for ( int i = 0; i < x; i++)
num = (num ^ (1 << i));
cout << num;
}
int main()
{
int num = 11;
invertBits(num);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static void invertBits( int num)
{
int x = ( int )(Math.log(num) /
Math.log( 2 )) + 1 ;
for ( int i = 0 ; i < x; i++)
num = (num ^ ( 1 << i));
System.out.println(num);
}
public static void main (String[] args)
{
int num = 11 ;
invertBits(num);
}
}
|
Python3
import math
def invertBits(num):
x = int (math.log2(num)) + 1
for i in range (x):
num = (num ^ ( 1 << i))
print (num)
num = 11
invertBits(num)
|
C#
using System;
class GFG
{
static void invertBits( int num)
{
int x = ( int )(Math.Log(num) /
Math.Log(2)) + 1;
for ( int i = 0; i < x; i++)
num = (num ^ (1 << i));
Console.WriteLine(num);
}
public static void Main ()
{
int num = 11;
invertBits(num);
}
}
|
Javascript
<script>
function invertBits(num)
{
let x =
parseInt(Math.log(num) / Math.log(2), 10) + 1;
for (let i = 0; i < x; i++)
num = (num ^ (1 << i));
document.write(num);
}
let num = 11;
invertBits(num);
</script>
|
PHP
<?php
function invertBits( $num )
{
$x = log( $num ) + 1;
for ( $i = 0; $i < $x ; $i ++)
$num = ( $num ^ (1 << $i ));
echo $num ;
}
$num = 11;
invertBits( $num );
?>
|
Time complexity : O(log n)
Auxiliary space : O(1)
Method 2 (Using bit-wise operators)
Think about generating a number just less than or equal to ‘n’ with all bits ‘1’ and take xor of that number with ‘n’.
Algorithm:
1. A variable ‘x’ is initialized to 1.
2. The function enters a while loop that continues until ‘x’ is less than or equal to the input integer (A).
3. Inside the while loop, ‘x’ is left-shifted by 1 using the bit-wise left shift operator (<<=). This multiplies ‘x’ by 2.
4. After the while loop, the value of ‘x’ is decremented by 1.
5. The function returns the result of the bit-wise XOR operation between ‘x’ and the input integer (A)
6. The final result is the bit-wise XOR of the input integer with the largest power of 2 that is less than or equal to the input integer.
Below is the implementation of the above idea:
C++
#include <bits/stdc++.h>
using namespace std;
class GFG {
public :
int Opposite( int n)
{
int x = 1;
while (x <= n) {
x <<= 1;
}
x--;
return (x ^ n);
}
};
int main()
{
int n = 5;
GFG Obj = GFG();
cout << Obj.Opposite(n);
return 0;
}
|
Java
import java.util.*;
class GFG {
public int Opposite( int n) {
int x = 1 ;
while (x <= n){
x <<= 1 ;
}
x--;
return (x ^ n);
}
}
public class Main {
public static void main(String[] args) {
int n = 5 ;
GFG Obj = new GFG();
System.out.print(Obj.Opposite(n));
}
}
|
Python3
class GFG:
def Opposite( self , n):
x = 1
while x < = n:
x << = 1
x - = 1
return (x ^ n)
n = 5
Obj = GFG()
print (Obj.Opposite(n))
|
C#
using System;
public class GFG {
public int Opposite( int n)
{
int x = 1;
while (x <= n) {
x <<= 1;
}
x--;
return (x ^ n);
}
}
class Program {
static void Main( string [] args)
{
int n = 5;
GFG Obj = new GFG();
Console.WriteLine(Obj.Opposite(n));
}
}
|
Javascript
class GFG {
Opposite(n) {
let x = 1;
while (x <= n) {
x <<= 1;
}
x--;
return (x ^ n);
}
}
let n = 5;
let Obj = new GFG();
console.log(Obj.Opposite(n));
|
Time Complexity: O(log n)
Auxiliary Space: O(1)
Method 3 (Using Bitset)
Here we use the flip() of bitset to invert the bits of the number, in order to avoid flipping the leading zeroes in the binary representation of the number, we have calculated the number of bits in the binary representation and flipped only the actual bits of the number. We have used to_ulong() to convert bitset to number.
C++
#include <bits/stdc++.h>
using namespace std;
void invertBits( int num)
{
int x = log2(num) + 1;
bitset<32> b(num);
for ( int i = 0; i < x; i++)
b.flip(i);
cout << b.to_ulong();
}
int main()
{
int num = 11;
invertBits(num);
return 0;
}
|
Java
class GFG
{
static void invertBits( int num)
{
int x = ( int )(Math.log(num) /
Math.log( 2 )) + 1 ;
for ( int i = 0 ; i < x; i++)
num = (num ^ ( 1 << i));
System.out.print(num);
}
public static void main(String[] args)
{
int num = 11 ;
invertBits(num);
}
}
|
Python 3
import math
def invertBits(num):
x = int (math.log(num, 2.0 ) + 1 );
for i in range ( 0 , x):
num = (num ^ ( 1 << i));
print (num);
num = 11 ;
invertBits(num);
|
C#
using System;
class GFG
{
static void invertBits( int num)
{
int x = ( int )Math.Log(num, 2) + 1;
for ( int i = 0; i < x; i++)
num = (num ^ (1 << i));
Console.Write(num);
}
static void Main()
{
int num = 11;
invertBits(num);
}
}
|
Javascript
<script>
function invertBits(num)
{
let x = Math.log(num, 2) + 1;
for (let i = 0; i < x; i++)
num = (num ^ (1 << i));
document.write(num);
}
let num = 11;
invertBits(num);
</script>
|
PHP
<?php
function invertBits( $num )
{
$x = log( $num ) + 1;
for ( $i = 0; $i < $x ; $i ++)
$num = ( $num ^ (1 << $i ));
echo $num ;
}
$num = 11;
invertBits( $num );
?>
|
Time complexity: O(log n)
Auxiliary space: O(1)
Method 4 (Using bitmask):
Here, we will create a mask by setting all the bits from MSB(including MSB) and then take XOR with the original number
C++
#include <bits/stdc++.h>
using namespace std;
void invertBits( int num)
{
int x = num;
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
cout << (num ^ x);
}
int main()
{
int num = 11;
invertBits(num);
return 0;
}
|
Java
import java.io.*;
class GFG {
public static void invertBits( int num) {
if (num == 0 )
{
System.out.println( 1 );
return ;
}
int x = num;
x |= x >> 1 ;
x |= x >> 2 ;
x |= x >> 4 ;
x |= x >> 8 ;
x |= x >> 16 ;
System.out.println(num ^ x);
}
public static void main(String[] args)
{
int num = 11 ;
invertBits(num);
}
}
|
Python3
def invertBits(num):
x = num
x | = x >> 1
x | = x >> 2
x | = x >> 4
x | = x >> 8
x | = x >> 16
print (num ^ x)
num = 11
invertBits(num)
|
C#
using System;
class GFG {
public static void invertBits( int num)
{
if (num == 0) {
Console.WriteLine(1);
return ;
}
int x = num;
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
Console.WriteLine(num
^ x);
}
public static void Main( string [] args)
{
int num = 11;
invertBits(num);
}
}
|
Javascript
function invertBits(num)
{
let x = num;
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
console.log(num ^ x);
}
let num = 11;
invertBits(num);
|
Time complexity: O(1)
Auxiliary space: O(1)
Method 5 (Extracting only the relevant bits using log and XOR)
The inverted number can be efficiently obtained by:
1. Getting the number of bits using log2
2. Taking XOR of the number and 2 numOfBits – 1
C++
#include <bits/stdc++.h>
using namespace std;
void invertBits( int num)
{
int numOfBits = ( int )log2(num) + 1;
cout << (((1 << numOfBits) - 1) ^ num);
}
int main()
{
int num = 11;
invertBits(num);
return 0;
}
|
Java
import java.io.*;
class GFG {
static void invertBits( int num)
{
int numOfBits = ( int )(Math.log(num) / Math.log( 2 )) + 1 ;
System.out.println((( 1 << numOfBits) - 1 ) ^ num);
}
public static void main(String[] args)
{
int num = 11 ;
invertBits(num);
}
}
|
Python
def invertBits(num):
numOfBits = num.bit_length()
print ((( 1 << numOfBits) - 1 ) ^ num)
num = 11
invertBits(num)
|
C#
using System;
class GFG
{
static void invertBits( int num)
{
int numOfBits = ( int )(Math.Log(num) / Math.Log(2)) + 1;
Console.WriteLine(((1 << numOfBits) - 1) ^ num);
}
public static void Main( string [] args)
{
int num = 11;
invertBits(num);
}
}
|
Javascript
function invertBits(num) {
let numOfBits = Math.floor(Math.log2(num)) + 1;
console.log(((1 << numOfBits) - 1) ^ num);
}
let num = 11;
invertBits(num);
|
Time complexity: O(1)
Auxiliary space: O(1)
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